Algebra... or Geometry?!

Geometry Level 4

Given that positive real numbers x,y and z satisfy following conditions.

x 2 + x y + y 2 3 = 25 x^{2}+xy+\frac{y^{2}}{3}=25

y 2 3 + z 2 = 9 \frac{y^{2}}{3}+z^{2}=9

z 2 + z x + x 2 = 16 z^{2}+zx+x^{2}=16

Find the value of 3 \sqrt{3} (xy+2yz+3zx)


The answer is 72.

Lee Young Kyu by Lee Young Kyu , 25, Hong Kong

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1 solution

Lee Young Kyu
Aug 4, 2014

Try to use cosine rule and you will see the three triangles will form a large right angled triangle.... And you try to express the area of the large triangle in form of x,y and z and you can easily find the answer.

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The second equation should be y 2 3 + z 2 = 9 \frac{y^2}{3} + z^2 = 9 , not y 3 3 + z 2 = 9 \frac{y^3}{3} + z^2 = 9 .

Jon Haussmann - 6 years, 10 months ago

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Oops I've editted it, thank you very much!

Lee Young Kyu - 6 years, 10 months ago

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